3.1.3 \(\int x \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [3]

Optimal. Leaf size=75 \[ -\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 e}+\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

1/4*d*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/e+1/2*x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-1/4*x*(e*x^2+d)^(1/2)/e^
(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6356, 327, 223, 212} \begin {gather*} -\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {d \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

-1/4*(x*Sqrt[d + e*x^2])/Sqrt[e] + (d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(4*e) + (x^2*ArcTanh[(Sqrt[e]*x)/S
qrt[d + e*x^2]])/2

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 6356

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(ArcT
anh[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x] - Dist[c/(d*(m + 1)), Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /;
FreeQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{2} \sqrt {e} \int \frac {x^2}{\sqrt {d+e x^2}} \, dx\\ &=-\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {d \int \frac {1}{\sqrt {d+e x^2}} \, dx}{4 \sqrt {e}}\\ &=-\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {d \text {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{4 \sqrt {e}}\\ &=-\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 e}+\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 76, normalized size = 1.01 \begin {gather*} -\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {1}{2} x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {d \log \left (\sqrt {e} x+\sqrt {d+e x^2}\right )}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

-1/4*(x*Sqrt[d + e*x^2])/Sqrt[e] + (x^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/2 + (d*Log[Sqrt[e]*x + Sqrt[d +
e*x^2]])/(4*e)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(156\) vs. \(2(57)=114\).
time = 0.01, size = 157, normalized size = 2.09

method	result	size
default	\(\frac {x^{2} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{2}+\frac {e^{\frac {3}{2}} \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{2 d}-\frac {\sqrt {e}\, \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 d}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/2*e^(3/2)/d*(1/4*x^3/e*(e*x^2+d)^(1/2)-3/4*d/e*(1/2*x/e*(e*x^2+d)
^(1/2)-1/2*d/e^(3/2)*ln(x*e^(1/2)+(e*x^2+d)^(1/2))))-1/2*e^(1/2)/d*(1/4*x*(e*x^2+d)^(3/2)/e-1/4*d/e*(1/2*x*(e*
x^2+d)^(1/2)+1/2/e^(1/2)*d*ln(x*e^(1/2)+(e*x^2+d)^(1/2))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/4*x^2*log(x*e^(1/2) + sqrt(x^2*e + d)) - 1/4*x^2*log(-x*e^(1/2) + sqrt(x^2*e + d)) - 1/2*d*integrate(-x^2*e^
(1/2*log(x^2*e + d) + 1/2)/(x^4*e^2 + d*x^2*e - (x^2*e + d)^2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (56) = 112\).
time = 0.35, size = 168, normalized size = 2.24 \begin {gather*} \frac {{\left (2 \, x^{2} \cosh \left (\frac {1}{2}\right )^{2} + 4 \, x^{2} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + 2 \, x^{2} \sinh \left (\frac {1}{2}\right )^{2} + d\right )} \log \left (\frac {2 \, x^{2} \cosh \left (\frac {1}{2}\right )^{2} + 4 \, x^{2} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + 2 \, x^{2} \sinh \left (\frac {1}{2}\right )^{2} + 2 \, {\left (x \cosh \left (\frac {1}{2}\right ) + x \sinh \left (\frac {1}{2}\right )\right )} \sqrt {\frac {{\left (x^{2} + d\right )} \cosh \left (\frac {1}{2}\right ) + {\left (x^{2} - d\right )} \sinh \left (\frac {1}{2}\right )}{\cosh \left (\frac {1}{2}\right ) - \sinh \left (\frac {1}{2}\right )}} + d}{d}\right ) - 2 \, {\left (x \cosh \left (\frac {1}{2}\right ) + x \sinh \left (\frac {1}{2}\right )\right )} \sqrt {\frac {{\left (x^{2} + d\right )} \cosh \left (\frac {1}{2}\right ) + {\left (x^{2} - d\right )} \sinh \left (\frac {1}{2}\right )}{\cosh \left (\frac {1}{2}\right ) - \sinh \left (\frac {1}{2}\right )}}}{8 \, {\left (\cosh \left (\frac {1}{2}\right )^{2} + 2 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + \sinh \left (\frac {1}{2}\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/8*((2*x^2*cosh(1/2)^2 + 4*x^2*cosh(1/2)*sinh(1/2) + 2*x^2*sinh(1/2)^2 + d)*log((2*x^2*cosh(1/2)^2 + 4*x^2*co
sh(1/2)*sinh(1/2) + 2*x^2*sinh(1/2)^2 + 2*(x*cosh(1/2) + x*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*si
nh(1/2))/(cosh(1/2) - sinh(1/2))) + d)/d) - 2*(x*cosh(1/2) + x*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d
)*sinh(1/2))/(cosh(1/2) - sinh(1/2))))/(cosh(1/2)^2 + 2*cosh(1/2)*sinh(1/2) + sinh(1/2)^2)

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Sympy [A]
time = 0.33, size = 66, normalized size = 0.88 \begin {gather*} \begin {cases} \frac {d \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{4 e} + \frac {x^{2} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{2} - \frac {x \sqrt {d + e x^{2}}}{4 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((d*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(4*e) + x**2*atanh(sqrt(e)*x/sqrt(d + e*x**2))/2 - x*sqrt(d + e
*x**2)/(4*sqrt(e)), Ne(e, 0)), (0, True))

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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